Решите неравенство: (x^2+x)^2-8*(x^2+x)+12<=0.

\[t = x^{2} + x\]

\[t^{2} - 8t + 12 \leq 0\]

\[(t - 6)(t - 2) \leq 0\]

\[2 \leq t \leq 6 \Longrightarrow 2 \leq x^{2} + x \leq 6\]

\[\left\{ \begin{matrix} x^{2} + x \geq 2 \\ x^{2} + x \leq 6 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2} + x - 2 \geq 0 \\ x^{2} + x - 6 \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} (x - 1)(x + 2) \geq 0 \\ (x - 2)(x + 3) \leq 0 \\ \end{matrix} \right.\ \]

\[Ответ:\lbrack - 3; - 2\rbrack \cup \lbrack 1;2\rbrack.\]